SOLUTION: Simplify : (6-a-a^2)/(a^2-4) So I factor to cancel terms (2-a)(3+a)/(a+2)(a-2) And so I get (3+a)/(a+2) BUT The answer on the book is (-3+a)/(a+2) Where did the n

Algebra ->  Equations -> SOLUTION: Simplify : (6-a-a^2)/(a^2-4) So I factor to cancel terms (2-a)(3+a)/(a+2)(a-2) And so I get (3+a)/(a+2) BUT The answer on the book is (-3+a)/(a+2) Where did the n      Log On


   

Question 1206053: Simplify :
(6-a-a^2)/(a^2-4)
So I factor to cancel terms
(2-a)(3+a)/(a+2)(a-2)
And so I get
(3+a)/(a+2)
BUT
The answer on the book is
(-3+a)/(a+2)
Where did the negative sign come from ?
Please help.thanks In advance !
Found 2 solutions by mananth, greenestamps:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
(6-a-a^2)/(a^2-4)

(6-3a+2a-a^2)/(a+2)(a-2)

(3(2-a)+a(2-a))/(a+2)(a-2)

(3+a)(2-a)/(a+2)(a-2)
Bring the negative sign out and change sign of (2-a) to (a-2)

-(3+a)(a-2)/(a+2)(a-2) Cancel off (a-2)

-(3+a)/(a+2)





Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


I hope you mean the answer in the book is -%283%2Ba%29%2F%28a%2B2%29 and not %28-3%2Ba%29%2F%28a%2B2%29....

In the work you show, you have....

%28%282-a%29%283%2Ba%29%29%2F%28%28a%2B2%29%28a-2%29%29

and then

%283%2Ba%29%2F%28a%2B2%29

That means you canceled the factors %282-a%29 and %28a-2%29.

But those factor are not the same; they are opposites: %282-a%29=-1%28a-2%29

So